Will splitting RCA's drop the HU voltage output?

bangandow
10+ year member

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I decided to completely nix the rear fill and use all 4 channels of my amp to run my front components... so I got 2 Y-adapters for the RCA's and bridged the amp into 2 channels...

Now, it sounds great, but in order to get any good amount of volume out of the system, I have to turn the gains up pretty **** high (like 70%+), whereas the headunit preouts are supposed to be 5volt...

I'm wondering if, when I split the RCA's, it halved the voltage each was seeing... does that make sense?

If not that, then what can account for the loss in volume? It seems like every couple weeks I have to try and tweek it to keep it at the volume I like, so I assumed I was going deaf, but then other people started noticing the same thing... so somewhere along the line, I'm steadily losing volume.

I know I got into a separate issue there at the end... sorry. If you can shed any light on either problem, it'd be much appreciated //content.invisioncic.com/y282845/emoticons/biggrin.gif.d71a5d36fcbab170f2364c9f2e3946cb.gif

 
I'm wondering if, when I split the RCA's, it halved the voltage each was seeing...
No, using y-splitters does not "split" the voltage.

I know I got into a separate issue there at the end... sorry. If you can shed any light on either problem, it'd be much appreciated //content.invisioncic.com/y282845/emoticons/biggrin.gif.d71a5d36fcbab170f2364c9f2e3946cb.gif
What does your system consist of?

Because you apparently have a separate issue. It has nothing to do with the y-splitters.

 
Use this calculator to figure it out.

http://www.sengpielaudio.com/calculator-voltagedivider.htm

Vin = Enter HU output voltage

Z1 = Enter HU output impedance

Z2 = Enter Amplifier #1 input impedance

Zload = Enter Amplifier #2 input impedance.

Example.

Vin = 5 volts

Z1 = 200 ohms

Z2 = 20,000 ohms

Zload = Don't enter anything, default is 1 megaohm.

The HU outputs 5V, but with amplifier #1 connected, the loading effect of the amp causes

the voltage to be 4.95 volts.

If you connect another amp, lets say it also has 20,000 ohms input impedance,

the calculator says;

Vout = 4.90 volts

With one amp, 4.95v

With two amps, 4.90v

*What if*

Vin = 5 volts

Z1 = 2000 ohms

Z2 = 10,000 ohms

Zload = 10,000 ohms

With two amps, 4.54v

 
it would split the signal in 1/2 though, just like splitting your Coax cable for your TV reduces the signal by 50%, wether or not it is noticable is another story.
There is no "signal splitting" when using y-splitters for audio.

It's basic parallel wiring - voltage remains the same at each circuit.

Relate it to your car's battery. Is the voltage cut in half each time you add a circuit to your battery? Or does each item you add still see the full 12V?

 
whoa whoa whoa......I clearly remeber when I split my RCA I had to turn the gain up a lot more to get the same power out of the amp. Why was this?

 
Vout = 4.90 volts

With one amp, 4.95v

With two amps, 4.90v
Yes, you will probably get some very, very, very minor voltage drop. Completely unnoticeable performance wise, unless you were stringing together a large multitude of amps.

But the signal isn't "split in half"....so each amplifier input (in the above example including the very minor hypothetical voltage drop) would see the whole 4.9V

 
It's basic parallel wiring - voltage remains the same at each circuit.
nope //content.invisioncic.com/y282845/emoticons/smile.gif.1ebc41e1811405b213edfc4622c41e27.gif
Yes, I was "glossing over" minor technical information.

The voltage each circuit sees is identical to each other. It's not "split in half" between them. This is what I was attempting to state.

But yes Mr. Technicality, due to various factors there is a minute amount of voltage drop.

 
*Secrets Revealed*

Lets take a typical Alpine HU. Lets pick this one.

CDA-9856

http://www.alpine-usa.com/US-en/products/product.php?model=CDA-9856&lang=en&tab=F

Notice how they don't list the output impedance. I wonder why ?

Lets download the manual. Click.

Page 28

Preout: 2 volts / 10,000 ohm output impedance..

Whoah............. //content.invisioncic.com/y282845/emoticons/mad.gif.c18f003ab0ef8a0d9c27ca78d77a6392.gif//content.invisioncic.com/y282845/emoticons/thumbsdown.gif.d22f25895e9b40f2300c953691dacfa2.gif//content.invisioncic.com/y282845/emoticons/plthumbsdown.gif.d7966af824c93c6021d1aae1778b1c19.gif

Connect this HU to an Alpine amp with 10,000 ohm input impedance,

the voltage is now 1 volt, cut in half.

//content.invisioncic.com/y282845/emoticons/bsflag.gif.21f42eccd34b7d1eb1608fb1b59b69c3.gif

Instead of buying an Alpine amp, you found a good amp with a 47,000 ohm

input impedance, the loading effect is minimal, you get 1.95v, much better

than 1.0v when you interfaced an Alpine HU to an Alpine amp.

I was shocked years ago when I found this.

//content.invisioncic.com/y282845/emoticons/laugh.gif.48439b2acf2cfca21620f01e7f77d1e4.gif

Connect two Alpine amps with splitter, you now have 0.66v.

Hint: A good HU will have a 50 ohm - 200 ohm output impedance. Even 1000 ohm is ok,

but 10,000 ohm is stupid circuit design.

Moral of story: Always check equipment specs.

 
I've "heard" that they meant you'll have 2V of output driving a 10k ohm load.

That said; I've not measured it myself. Then again, neither have you //content.invisioncic.com/y282845/emoticons/biggrin.gif.d71a5d36fcbab170f2364c9f2e3946cb.gif

 
I've "heard" that they meant you'll have 2V of output driving a 10k ohm load.
That said; I've not measured it myself. Then again, neither have you //content.invisioncic.com/y282845/emoticons/biggrin.gif.d71a5d36fcbab170f2364c9f2e3946cb.gif
When in doubt, assume the worst case or get burned. //content.invisioncic.com/y282845/emoticons/laugh.gif.48439b2acf2cfca21620f01e7f77d1e4.gif

They claim;

Preout: 2V / 10k ohm.

Lets look at Eclipse

They claim;

Output Voltage and Impedance: 8V / 55 ohm

http://www.eclipse-web.com/cd/feature-cd753.html

I interpret the spec as output impedance because another manufacturer

uses the same format, but unless you test it, there is doubt.

 
There is no "signal splitting" when using y-splitters for audio.
It's basic parallel wiring - voltage remains the same at each circuit.

Relate it to your car's battery. Is the voltage cut in half each time you add a circuit to your battery? Or does each item you add still see the full 12V?
I always Thought Voltage and signal were two different things....Just as voltage and current are related but completely different...

When you split a cable off your cars battery the VOLTAGE is not changed but the maximum Current through the split cable is (which can affect voltage since it is coming from a battery). If you split some 0ga wire into 2, 0ga wires then your limiting the current at the split, you will be limited to the current of the original 0ga wire, so the wire after the split may as well be smaller gauge since they will only have 1/2 their typical current capabilities. Since you can run 30ga wire to your amp and get 14.8V to it without issues does not mean the amp is receiving enough power since you would horrifically be limiting the current.

Now I know for a fact TV signals degrade quite quickly when splitting the cables. Go to best buy, and look at their TV's that are all split off a single set of cables (In Parallel) some of them may look like *** even though they are quite good. This is because the signal is split too many times. You can't judge the TV set by the way it looks at Best Buy.

So I guess I don't understand how RCA's would be any different. Are they Magic? I thought audio signals were sine waves? Again the voltage drop may not be significant, but the signal strength itself after the load can't be as strong can it?

Please explain how this is any different.

thylantyr - what math are you doing? I'm not following why an amp with more input impedance would reduce the HU voltage more than an amp with less input impedance?

 
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